Difference between revisions of "009C Sample Final 1, Problem 2"

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!Step 2:  
 
!Step 2:  
 
|-
 
|-
−
|Since <math>2<e</math>, <math>\bigg|-\frac{2}{e}\bigg|<1</math>. So,  
+
|Since <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1</math>. So,  
 
|-
 
|-
−
|<math>\sum_{n=0}^{\infty} (-2)^ne^{-n}=\frac{1}{1+\frac{2}{e}}=\frac{1}{\frac{e+2}{e}}=\frac{e}{e+2}</math>.
+
|
 +
::<math>\sum_{n=0}^{\infty} (-2)^ne^{-n}=\frac{1}{1+\frac{2}{e}}=\frac{1}{\frac{e+2}{e}}=\frac{e}{e+2}</math>.
 
|}
 
|}
  
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|This is a telescoping series. First, we find the partial sum of this series.
 
|This is a telescoping series. First, we find the partial sum of this series.
 
|-
 
|-
−
|Let <math>s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>.
+
|Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>.
 
|-
 
|-
−
|Then, <math>s_k=\frac{1}{2}-\frac{1}{2^{k+1}}</math>.
+
|Then, <math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}</math>.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Thus, <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)=\lim_{k\rightarrow \infty} s_k=\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}=\frac{1}{2}</math>
+
|Thus,  
 +
|-
 +
|
 +
::<math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)=\lim_{k\rightarrow \infty} s_k=\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}=\frac{1}{2}</math>
 
|}
 
|}
  

Revision as of 13:23, 22 February 2016

Find the sum of the following series:

a)

b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)}

Foundations:  
Review geometric series.
Review telescoping series.

Solution:

(a)

Step 1:  
First, we write
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\sum_{n=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \bigg(\frac{-2}{e}\bigg)^n}\\ \end{array}}
Step 2:  
Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2<e,~\bigg|-\frac{2}{e}\bigg|<1} . So,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} (-2)^ne^{-n}=\frac{1}{1+\frac{2}{e}}=\frac{1}{\frac{e+2}{e}}=\frac{e}{e+2}} .

(b)

Step 1:  
This is a telescoping series. First, we find the partial sum of this series.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)} .
Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k=\frac{1}{2}-\frac{1}{2^{k+1}}} .
Step 2:  
Thus,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)=\lim_{k\rightarrow \infty} s_k=\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}=\frac{1}{2}}


Final Answer:  
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{e}{e+2}}
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}}

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